Problem: Find all values of $k$ so that the domain of
\[b(x) = \frac{kx^2 + 2x - 5}{-5x^2 + 2x + k}\]is the set of all real numbers.
Explanation: The domain of the function is the set of all real numbers if and only if the denominator $-5x^2 + 2x + k$ is nonzero for all $x.$  In other words, the quadratic
\[-5x^2 + 2x + k = 0\]should not have any real solutions.  This means that the discriminant is negative, i.e.
\[4 - 4(-5)(k) = 4 + 20k < 0.\]Solving, we find $k < -\frac{1}{5}.$  Therefore, the set of all possible $k$ is $\boxed{\left( -\infty, -\frac{1}{5} \right)}.$